Question

Consider the reaction A ⟶Products

Answer 1

The given reaction is A---> products

Trial-1: The initial concentration of A is 0.10 M when the initial rate is 0.015 M/s

Trial-2: The initial concentration of  A is 0.40 M when the initial rate is 0.060 M/s

Let the order be m

Rate law for first trial will be: $0.015M/s = k.[0.10M]^{m}$

Rate law for second trial will be:  $0.060M/s = k.[0.40M]^{m}$

Rate- 2/ Rate-1 : $\frac{0.060M/s}{0.0150M/s}=\frac{k(0.40)^{m}}{k.(0.10)^{m}}$

$4 = 4^{m}$

So, m = 1

Therefore the order of the reaction is 1